By [1]} the Minkowski sum of a compact, convex set and a closed, convex cone are closed since they do not have a common direction of recession. Thus, since \({\mathcal {F}}\)  is compact, convex and \({\mathcal {R}}_{ecc}\)  is a subspace and therefore a closed, convex cone, the set \({\mathcal {F}}+{\mathcal {R}}_{ecc}\)  is closed. Note that the sets
\(\hspace{-28.45274pt}\begin{split}&\left\lbrace \bar{{\mathsf {X}}}\in {\mathcal {S}}^{n_1+Sn_2} \colon \begin{pmatrix}{\mathsf {X}}& {\mathsf {Z}}_i{\mathsf {Z}}_i & {\mathsf {Y}}_{i,i}\end{pmatrix} \in {\mathcal {R}}_i \right\rbrace \\=&\left\lbrace \bar{{\mathsf {X}}}\in {\mathcal {S}}^{n_1+Sn_2} \colon \begin{pmatrix}{\mathsf {X}}& {\mathsf {Z}}_i{\mathsf {Z}}_i & {\mathsf {Y}}_{i,i}\end{pmatrix} \in {\mathcal {R}}_i,\ \mbox{all other entries are zero}  \right\rbrace \\&+\left\lbrace \bar{{\mathsf {X}}}\in {\mathcal {S}}^{n_1+Sn_2} \colon \begin{pmatrix}{\mathsf {X}}& {\mathsf {Z}}_i{\mathsf {Z}}_i & {\mathsf {Y}}_{i,i}\end{pmatrix} = {\mathsf {O}} \right\rbrace , \ i \in [1\!:\!S],\end{split} \) 
